Câu 1 :

a) \(CuO+2HCl\rightarrow CuCl_2+H_2O\left(1\right)\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\)

b) Gọi số mol CuO, ZnO lần lượt là x, y

\(m_{hh}=m_{CuO}+m_{ZnO}\rightarrow80x+81y=12,1\left(+\right)\)

\(n_{HCl}=0,1.3=0,3\left(mol\right)\)

Theo (1): nHCl (1) = 2nCuO = 2x

Theo (2): nHCl (2) = 2nZnO = 2y

\(n_{HCl}=2x+2y=0,3\left(++\right)\)

Từ (+) và (++) \(\rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

\(\%m_{CuO}=\frac{0,05.80}{12,1}.100\%=33,06\%\)

\(\%m_{ZnO}=100\%-33,06\%=66,94\%\)

c) \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,05_____0,05____________________

\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

0,1______0,1______________________

\(n_{H2SO4}=0,05+0,1=0,15\left(mol\right)\)

\(m_{H2SO4}=0,15.98=14,7\left(g\right)\)

\(m_{dd_{H2SO4}}=\frac{14,7}{20\%}=73,5\left(g\right)\)

Câu 2:

a) Gọi số mol Al, Fe là x, y

mhh = mAl + mFe

\(\rightarrow27x+56y=5,5\left(1\right)\)

\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

x_____3x______x________3x/2

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

y____2y_______y________y

\(n_{H2}=\frac{3x}{2}+y=0,2\left(1\right)\)

Từ (1) và (2) \(\rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\%m_{Al}=\frac{0,1.27}{5,5}.100\%=49,09\%\)

\(\%m_{Fe}=100\%-49,09\%=50,91\%\)

b) \(n_{HCl_{pu}}=3.0,1+2.0,05=0,4\left(mol\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{dd_{HCl}}=14,6:14,6\%=100\%\)

mdd sau phản ứng = mKL + mdd HCl - mH2

\(=5,5+100-0,2.2=105,1\left(g\right)\)

Sau phản ứng dung dịch có: AlCl3 (0,1 mol); FeCl2 (0,05 mol)

\(C\%_{AlCl3}=\frac{0,1.133,5}{105,1}.100\%=12,7\%\)

\(C\%_{FeCl2}=\frac{0,05.127}{105,1}.100\%=6,04\%\)

bài ni mik lm rồi, tham khảo nhé

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)

Ta có : \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)

=> x=0,2 ; y=0,1

 \(\%m_{Al}=\dfrac{0,2.27}{11}.100==49,09\%\)

\(\%m_{Fe}=50,91\%\)

b) \(\Sigma n_{HCl}=3x+2y=0,8\left(mol\right)\)

=> \(V_{HCl}=\dfrac{0,8}{2}=0,4\left(lít\right)\)

c) \(CM_{AlCl_3}=\dfrac{0,2}{0,4}=0,5M\)

\(CM_{FeCl_2}=\dfrac{0,1}{0,4}=0,25M\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2.........0.4.........0.2......0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\Rightarrow m_{ZnO}=14.6-13=1.6\left(g\right)\)

\(\%Zn=\dfrac{13}{14.6}\cdot100\%=89.04\%\)

\(\%ZnO=100\%-89.04\%=10.96\%\)

\(n_{ZnO}=\dfrac{1.6}{81}\approx0.02\left(mol\right)\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(0.02........0.04........0.02........0.02\)

\(n_{HCl}=0.4+0.04=0.44\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.44}{0.8}=0.55\left(M\right)\)

Eeeee ngồi tính sang chấn thật nó ra số xấu lần mò hơn 20p chưa biết tính sai chỗ nào

Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 5,5 (1)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,5}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

a)

Gọi : \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)⇒ 27a + 56b = 1,66(1)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe +2 HCl \to FeCl_2 + H_2\)

Theo PTHH :

\(n_{H_2} = 1,5a + b = \dfrac{1,12}{22,4} = 0,05(2)\)

Từ (1)(2) suy ra a = 0,02 ; b = 0,02

Vậy :

\(\%m_{Al} = \dfrac{0,02.27}{1,66}.100\% = 32,53\%\\ \%m_{Fe} = 100\% - 32,53\% = 67,47\%\)

a)

\(n_{HCl} = 2n_{H_2} = 0,05.2 = 0,1(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,1.36,5}{100}.100\% = 3,65\%\)

Gọi x,y lần lượt là số mol của Al, Fe

n_H2 = \(\dfrac{8,96}{22,4}\)=0,4 mol

Pt: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

......x.................................0,5x...........1,5x

.....Fe + H₂SO₄ --> FeSO₄ + H₂

.......y..........................y............y

Ta có hệ pt:

 {27x+56y=11

   1,5x+y=0,4

⇔x=0,2, y=0,1

% m_Al = \(\dfrac{0,2.27}{11}\).100%=49,1%

% m_Fe = \(\dfrac{0,1.56}{11}\).100%=50,9%

m_Al2(SO4)3 = 0,5x . 342 = 0,5 . 0,2 . 342 = 34,2 (g)

m_FeSO4 = 152y = 152 . 0,1 = 15,2 (g)

Gọi CTTQ: M_xO_y

Pt: M_xO_y + yH₂ --to--> xM + yH₂O

\(\dfrac{0,4}{y}\)<-------0,4

Ta có: 232,2=\(\dfrac{0,4}{y}\)(56x+16y)

⇔23,2=\(\dfrac{22,4x}{y}\)+6,4

⇔\(\dfrac{22,4x}{y}\)=16,8

⇔22,4x=16,8y

⇔x:y=3:4

Vậy CTHH của oxit: Fe₃O₄